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Factoring quartics-2 special cases!

part-1. i am going to show you some special of a quartic polynomial which makes it easier to solve. in future notes i will try to generalize for every quartic polynomials. 2 cases to be presented here:

case 1: consider the polynomial
\[f(x)=x^4+ax^3+bx^2+acx+c^2\]
i think most of you know how to factor this easily, but i am going to continue regardless.
\[\begin{array}a
f(x)=x^2(x^2+ax+b+acx^{-1}+c^2x^{-2})\\
\rightarrow x^2+2+(cx^{-1})^2+a(x+cx^{-1})+b-2=0\\
(x+cx^{-1})^2+a(x+cx^{-1})+b-2=0\\
x+cx^{-1}=\dfrac{-a\pm\sqrt{a^2-4b+8}}{2}\\
x^2-\dfrac{-a\pm\sqrt{a^2-4b+8}}{2}x+c=0\end{array}\] i think this is the most general form, as going farther will simply make it more tedious.
\[\]
case 2
\[x^4-2ax^2-x+a^2-a=0\]
\[x^4-2ax^2+a^2=a+x\]
\[x^2-a=\pm\sqrt{a+x}\]
\[x=\pm\sqrt{a+\sqrt{a+x}},\pm\sqrt{a-\sqrt{a+x}}\]
first one
\[x=\sqrt{a+\sqrt{a+x}}=\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+x}}}}=\sqrt{a+\sqrt{a+...}}=\sqrt{a+x}\]
\[x^2-x+a=0\]
the minus sign will have the negative root, and the positive sign the positive. second one:
\[x=\sqrt{a-\sqrt{a+\sqrt{a-\sqrt{a+...}}}},y=\sqrt{a+\sqrt{a-\sqrt{a+\sqrt{a-...}}}}\]
\[x^2=a-y,y^2=a+x\]
subtract both
\[x^2-y^2=-y-x\Longrightarrow x-y=-1\]
\[x^2=a-(x+1)\Longrightarrow x^2+x+(1-a)=0\]
we can actually simplify these two results to get our original polynomials.